Lecture 5 - 1st Law Unsteady Flow Flashcards
Pressure vessel in spanish
recipiente a presión
When dealing with an open system with unsteady flow, such as with a pressure vessel, we are dealing with a process that …
involves changes to the control volume with respect to time
If we have an open system with changes to the control volume with respect to time, we are dealing with …
unsteady flow
Charging a rigid vessel from a supply line, like a pressure vessel connected to a compressed airline, works as follows:
You open the valve, and the pressure vessel begins to charge
Discharging a pressurized vessel works as follows:
Water vapor is present through the compression process from the atmosphere, and when we quickly discharge a vessel, the water vapor in the air (around the nozzle of the vessel) gets very cold, and turns to liquid and, in some, cases ice.
We can use the first law of thermodynamics to solve problems involving unsteady flow by …
making approximations under the uniform flow process.
The uniform flow process
A specialized case where we assume
- the state of the entire control volume is uniform throughout
- the fluid flow at the inlet or exit is uniform and steady
- if properties do change during the process, we can find an average for the initial or final state of the inlet or outlet
The first law under the uniform flow process looks as follows:
Q - W = ∑mehe - ∑mihi + (m2u2 – m1u1)cv [kJ]
where Q = heat transfer W = work me = mass exiting mi = mass entering he = enthalpy of exiting fluid hi = enthalpy of entering fluid (m2u2 – m1u1)cv = change in internal energy of control volume m2 = mass at state 2 m1 = mass at state 1 [kJ] = units in kilojoules
The first law of thermodynamics with unsteady flow does not take into account kinetic or potential energy.
True
If we have 75% of X in liquid form, the quality of the system would be …
x = mass vapor/total mass x = 0.25/1
At the saturated vapor line we have …
all vapor and no liquid
In the first law of thermodynamics, if the system is not moving macroscopically (the entire system moves), and the system is not moving vertically with respect to gravity, we can approximate ∆KE and ∆PE to …
0 in both cases
The units for boundary work is …
kJ
The units for the equation W(boundary) = P∫dv is …
kJ
1∫2 dv =
V2 - V1
